A question for the smarter folk.
Does the type of plumbing in a building affect the standing losses/ heat losses during usage from a geyser.
Galv vs Copper vs PVC
Wouldnt really makes a difference if you insulated the pipes. That being said, I insulated all my copper pipes (all in-ceiling) and it made bugger all difference.
@calypso , when you say you insulated the copper pipes in the ceiling, do you mean the copper pipes in the attic ??
For standing losses the piping would not make a large difference as the hot water is not standing in the pipes but the geyser.
However when it comes to the usage of the hot water the piping does come into effect.
The distance from the geyser to where you use it would be the largest factor and then the pipe type will also have an impact (but smaller).
Any hot water standing in a pipe will cool down. So when you open your hot water tap all that cold water has to flow out before it is replaced with hot water from the geyser.
The longer the pipe to the tap the more cold water needs to be displaced.
Then the hot water in the pipe start cooling down again.
If you frequently use the hot water then it might be worth insulating the pipes, but if it just for evening and morning showers it would not be worth it.
The best option is still to have the pipe as short as possible.
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Yup. Some of it does sneak out via the first two meters or so of piping attached to the geyser (conduction basically), so if you insulate that part of the piping, the standing losses from the rest is insignificant.
Don’t know the precise math, but because heat travels at a rate proportional to the delta, the losses from the piping as a function of the distance away from the geyser will likely have the letter e in it, that is to say, for every meter the losses will be 2.7 times less 
regarding the cold water loss when opening a hot tap - during the drought there was a valve that you could buy that would allow you to save or divert that cold water that came out of the hot tap?
In my situation I can feed this cold water back to my jojo tanks for reuse.
can anyone recall this bypass valve? or where to purchase some?
found this online.
Rather set the house up with a hot water recirculating pump. These valves in the link above are expensive and it would be annoying to open the hot water tap and for nothing to come out until it’s hot. Guests would be confused AF.
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Interesting, I have never heard of hot water recirculating pumps.
@spiff this is similar to what you posted, as it is used in heat pump systems, and comes in in the same price range as a recirculating pump
I guess I need to just win the lotto and build a house with a whole bunch of “fancy stuff”
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Also @JacoDeJongh thanks for the phonecall, I think I understand some of the principle a little better, ill ask you to draw me a picture when you have time, hahaha
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Only a pleasure,
Will do, I think I have some Diagrams I can send you.
my geyser is located in the attic, horizontal position, e.g, last night, (4 degree ambient temperature,), the drop was about 13 degrees, no water was being used at that time of the night, is that normal
The other day after a long hot shower at 10 pm, the temperature dropped down to 26 degrees, but then it stayed at 26 the next morning.
is this due to the difference between geyser temperature and ambient temperature
Recirculating / looping hot water from my SG & GG sounds like a good plan, but based on my previous experience with plumbers - the problem is to find a plumber that understands this and can actually do the job?
It is hard to say, because it depends on how you measured, and so forth. For example, my heat pump has a probe measuring in the middle of the tank, so even if the LCD display says 35°C, it may be significantly above that, and it may also be significantly colder at the bottom of the tank.
But with regards to the quoted part: Yes. The water cools down at a rate proportional to the difference between the temperature of the water and ambient. It’s called Newton’s law of cooling.
Any time you have a rate that is proportional to what is already there, the formula will involve the mathematical constant e.
In the case of cooling, the formula is: T = ce^(kt). By substituting t=0 (the temperature at the beginning of cooling), we know c must be the temperature when it starts cooling down (usually 55°C or higher). Then we can plug in a new t (for some hours later) and the measured temperature, and solve for k. And then you have a formula for your own geyser
You can also use the standing loss spec for the geyser, convert that to an expected temperature after 24 hours, then plug in t=24 hours and the new calculated T, and solve for k. Then you have both c and k and you can plot the thing.
Also see this, if this is your sort of thing
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If you want to be extra fancy, you can do that same with your cold water running it through a chiller. Because who would want warm water coming out of your cold water tap on a hot summers day. That’s for peasants!
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